Finding function given limit
$begingroup$
$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.
I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?
limits
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
$endgroup$
add a comment |
$begingroup$
$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.
I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?
limits
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
$endgroup$
1
$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45
1
$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46
add a comment |
$begingroup$
$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.
I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?
limits
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
$endgroup$
$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.
I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?
limits
limits
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
edited Nov 23 '18 at 22:53
gimusi
92.8k84494
92.8k84494
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
asked Nov 23 '18 at 22:43
Amanda ChoiAmanda Choi
191
191
1
$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45
1
$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46
add a comment |
1
$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45
1
$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46
1
1
$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45
$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45
1
1
$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46
$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.
Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$
$frac{2+a}4=3$ ---> $a=10$
Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.
Then $c=-8, d=-20$.
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
$endgroup$
add a comment |
$begingroup$
We have that
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$
requires that for $x=2$
$$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$
then
$$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$
then
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$
As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$
from which we can find $c$ and then $d$.
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
$endgroup$
– Christopher Marley
Nov 23 '18 at 22:58
1
$begingroup$
@ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
$endgroup$
– gimusi
Nov 23 '18 at 23:01
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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votes
$begingroup$
Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.
Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$
$frac{2+a}4=3$ ---> $a=10$
Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.
Then $c=-8, d=-20$.
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
$endgroup$
add a comment |
$begingroup$
Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.
Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$
$frac{2+a}4=3$ ---> $a=10$
Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.
Then $c=-8, d=-20$.
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
$endgroup$
add a comment |
$begingroup$
Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.
Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$
$frac{2+a}4=3$ ---> $a=10$
Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.
Then $c=-8, d=-20$.
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
$endgroup$
Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.
Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$
$frac{2+a}4=3$ ---> $a=10$
Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.
Then $c=-8, d=-20$.
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
edited Nov 23 '18 at 22:58
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
answered Nov 23 '18 at 22:50
Christopher MarleyChristopher Marley
1,117115
1,117115
add a comment |
add a comment |
$begingroup$
We have that
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$
requires that for $x=2$
$$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$
then
$$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$
then
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$
As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$
from which we can find $c$ and then $d$.
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
$endgroup$
– Christopher Marley
Nov 23 '18 at 22:58
1
$begingroup$
@ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
$endgroup$
– gimusi
Nov 23 '18 at 23:01
add a comment |
$begingroup$
We have that
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$
requires that for $x=2$
$$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$
then
$$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$
then
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$
As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$
from which we can find $c$ and then $d$.
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
$endgroup$
– Christopher Marley
Nov 23 '18 at 22:58
1
$begingroup$
@ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
$endgroup$
– gimusi
Nov 23 '18 at 23:01
add a comment |
$begingroup$
We have that
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$
requires that for $x=2$
$$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$
then
$$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$
then
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$
As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$
from which we can find $c$ and then $d$.
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
$endgroup$
We have that
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$
requires that for $x=2$
$$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$
then
$$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$
then
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$
As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain
$$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$
from which we can find $c$ and then $d$.
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
edited Nov 23 '18 at 22:56
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
answered Nov 23 '18 at 22:49
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
$endgroup$
– Christopher Marley
Nov 23 '18 at 22:58
1
$begingroup$
@ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
$endgroup$
– gimusi
Nov 23 '18 at 23:01
add a comment |
$begingroup$
Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
$endgroup$
– Christopher Marley
Nov 23 '18 at 22:58
1
$begingroup$
@ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
$endgroup$
– gimusi
Nov 23 '18 at 23:01
$begingroup$
Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
$endgroup$
– Christopher Marley
Nov 23 '18 at 22:58
$begingroup$
Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
$endgroup$
– Christopher Marley
Nov 23 '18 at 22:58
1
1
$begingroup$
@ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
$endgroup$
– gimusi
Nov 23 '18 at 23:01
$begingroup$
@ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
$endgroup$
– gimusi
Nov 23 '18 at 23:01
add a comment |
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$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45
1
$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46